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3 years ago

(-2) + 4 = 4 + (-2)

Mathematics

2 answers:

Nataly_w [17]3 years ago

8 0

It's D the comunicative property of addition

g100num [7]3 years ago

3 0

Answer:

D: The Commutative Property of Addition

Step-by-step explanation:

This property states that changing the order of the numbers we are adding, does not change the sum.

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Austin earns $9.75 an hour at a part-time job, and earns $12 each time mowing lawns in the neighborhood. They graph the amount o

kaheart [24]

Answer:

$7.25 at part-time job

$9.00 every time lawn mowed

so he would get 33.33 day so we will just say 33 times to get 300 dollars if he gets $9 for mowing. And 41.37 hours to get $300 and he would have to use x and y graph to find the boundary line.

Step-by-step explanation:

4 0

2 years ago

15 less than a number tripled is equivalent to the difference of the same number and 12

Nitella [24]

Step-by-step explanation:

x = number

3x - 15 = (x - 12)×-4 or (12-x)×-4

3x - 15 = -4x + 48 or -48 + 4x

7x = 63, x = 9 or

33 = x

4 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

3 years ago

HELP PLS ASP WITH THE CORRECT ANSWER I WILL GUVE YOU 17 POINTS PLS ASP!!!! Drag and drop the angles pairs to correctly match the

Tems11 [23]

COB and COD complementary
AOB and BOC adjacent angle that is neither comp. or supp.
AOC and COD supplementary
AOE and COD vertical

6 0

3 years ago

Olivia has 48 slices of ham and 36 slices of cheese to make sandwiches.

serious [3.7K]

The greatest number of sandwiches that Olivia can make is 12.

3 slices of cheese will be on each sandwich

<u>Solution:</u>

Given that, Olivia has 48 slices of ham and 36 slices of cheese to make sandwiches.

We have to find what is the greatest number of sandwiches that Olivia can make? And how many slices of cheese will be on each sandwich?

Let the number of sandwiches be "n"

Then, number of slices of ham per sandwich $=\frac{48}{n}$

And, number of slices of cheese per sandwich $=\frac{36}{n}$

As slices can’t be fractions, "n" should be a factor of 48 and 36. So now let us find H.C.F of 48 and 36 as we want greatest number of sandwiches that Olivia can make.

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48

The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, and 36

The common factors are those which appear on both factors of 48 and 36 are 1, 2, 4, 6, and 12.

Out of these common factors, 12 is the greatest

So, H.C.F of 48 and 36 is 12, which means n = 12

Then, number of cheese slices per sandwich $=\frac{36}{12}=3$

Hence, 12 sandwiches can be made with 3 slices of cheese per sandwich.

8 0

3 years ago